In the oxygen sag calculation, what does the critical time t_c represent?

The time at which the oxygen deficit is greatest, i.e. DO is lowest

The time at which BOD is fully exhausted

The time at which the deficit becomes zero

The time at which reaeration stops

How are the critical deficit and minimum DO obtained in this answer?

D_c = (k1/k2) L0 e^(-k1 t_c); DO_min = DO_sat - D_c = 3.54 mg/L

D_c = (k2/k1) L0 e^(-k1 t_c); DO_min = D_c = 5.56 mg/L

D_c = D0 e^(-k2 t_c); DO_min = DO_sat + D_c

D_c = L0 - D0; DO_min = 9.1 - 2 = 7.1 mg/L

How is the distance of the minimum DO point found for a river flowing at 0.3 m/s?

Convert velocity to m/day (0.3 x 86 400) and multiply by t_c, giving about 76 km

Divide t_c by the velocity in m/day

Multiply t_c in days by 0.3 directly

Divide the critical deficit by the velocity

Practice · 8 marks

Critical DO deficit and location downstream

Short answer

Using Streeter‑Phelps: critical time = 2.94 days, critical deficit = 5.56 mg/L, so minimum DO = 9.1 – 5.56 = 3.54 mg/L. At velocity 0.3 m/s the critical point occurs 76 km downstream. After 1 day, deficit = 4.31 mg/L (DO = 4.79 mg/L).

A town discharges treated sewage into a river. After mixing, the BOD ultimate of the river water is L0 = 20 mg/L and the initial oxygen deficit is D0 = 2 mg/L. The saturation DO is 9.1 mg/L, deoxygenation constant k1 = 0.2 per day and reaeration constant k2 = 0.4 per day (base e, at the river temperature). The river flows at 0.3 m/s. Calculate (a) the critical time, (b) the critical deficit and the minimum DO, (c) the distance downstream at which the minimum DO occurs, (d) the DO deficit after 1 day.

Given: $L_0 = 20$ mg/L, $D_0 = 2$ mg/L, $DO_{sat} = 9.1$ mg/L, $k_1 = 0.2\ d^{-1}$ , $k_2 = 0.4\ d^{-1}$ , $u = 0.3$ m/s.

(a) Critical time

t_c = \frac{1}{k_2 - k_1}\ln\left[\frac{k_2}{k_1}\left(1 - \frac{D_0(k_2 - k_1)}{k_1 L_0}\right)\right]

\begin{aligned} \frac{D_0(k_2 - k_1)}{k_1 L_0} &= \frac{2\times0.2}{0.2\times20} = 0.1 \\ t_c &= \frac{1}{0.2}\ln\left[2\,(1 - 0.1)\right] = 5\ln(1.8) = 5\times0.5878 = 2.94\ \text{days} \end{aligned}

(b) Critical deficit and minimum DO

D_c = \frac{k_1}{k_2}L_0\,e^{-k_1 t_c} = \frac{0.2}{0.4}\times20\times e^{-0.2\times2.94} = 10\times e^{-0.5878} = 10\times0.5556 = 5.56\ \text{mg/L}

DO_{min} = DO_{sat} - D_c = 9.1 - 5.56 = 3.54\ \text{mg/L}

(c) Distance

Velocity $= 0.3\times86\,400 = 25\,920$ m/day.

x_c = 25\,920 \times 2.94 = 76\,200\ \text{m} \approx 76\ \text{km}

(d) Deficit after 1 day

\begin{aligned} D_1 &= \frac{0.2\times20}{0.4 - 0.2}\left(e^{-0.2} - e^{-0.4}\right) + 2\,e^{-0.4} \\ &= 20\,(0.8187 - 0.6703) + 2\times0.6703 = 2.968 + 1.341 = 4.31\ \text{mg/L} \end{aligned}

DO after 1 day $= 9.1 - 4.31 = 4.79$ mg/L.

Time (days)	0	1	2.94 (critical)
Deficit D (mg/L)	2.00	4.31	5.56
DO (mg/L)	7.10	4.79	3.54

Answer: $t_c$ = 2.94 days; $D_c$ = 5.56 mg/L; minimum DO = 3.54 mg/L; at about 76 km downstream; deficit after 1 day = 4.31 mg/L. The minimum DO is below the 4-5 mg/L usually needed for fish, so further treatment would be needed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Critical DO deficit and location downstream

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(d) Deficit after 1 day

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